# Evaluate indefinite integral $\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} dx$

How can one evaluate the indefinite integral $$\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} \,dx$$

First I tried to multiply by $$\frac{\sqrt{x}+1}{\sqrt{x}+1}$$, but I couldn't get the final result. I also tried to seperate the integral into two $$\int \frac{\sqrt{x}}{\sqrt{x}-1} dx + \int \frac{1}{\sqrt{x}-1} dx$$

which failed to lead to a solution either. Is there another method that can be used?

• What about the substitution $\sqrt x=t \Rightarrow x=t^2 \Rightarrow dx=2t\;dt$ ?
– User
Commented Oct 5, 2021 at 21:31
• @MasterShifu I did it to, thank you. Commented Oct 5, 2021 at 21:35
• You're welcome. By the way if you want to separate the fraction first, I would suggest writing it as $\frac{(\sqrt x-1)+2}{\sqrt x-1}=1+\frac2{\sqrt x-1}$.
– User
Commented Oct 5, 2021 at 21:47

Hint The form of the denominator suggests the substitution $$t = \sqrt{x} - 1 .$$ In particular, $$x$$ can be expressed as a polynomial in $$t$$, so the substitution yields a rational integrand with denominator $$t$$.
• @Student you have to take special care when $0 > t$, since then, $\log(t)$ is undefined. When $0 > t$, $\int \frac{dt}{t} = \int \frac{-dt}{-t} = - \log(-t).$ Commented Oct 5, 2021 at 22:43
\begin{align} &\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} \,dx\\ =& \int \frac{x+\sqrt{x}}{\sqrt x(\sqrt{x}-1)} \,dx =\int \frac{(\sqrt{x}+2)(\sqrt x-1)+2}{\sqrt x(\sqrt{x}-1)} \,dx\\ =& \int 1+\frac2{\sqrt x}+\frac{2}{\sqrt x(\sqrt{x}-1)} \,dx =x +4\sqrt x+4\ln|\sqrt x-1|+C \end{align}
\begin{aligned} \int \frac{\sqrt{x}+1}{\sqrt{x}-1} d x & =2 \int \frac{x+\sqrt{x}}{\sqrt{x}-1} d(\sqrt{x}) \\ & =2 \int \frac{y^2+y}{y-1} d y, \textrm{ where } y=\sqrt{x } \\ & =2 \int\left(y+2+\frac{2}{y-1}\right) d y \\ & =y^2+4 y+4 \ln |y-1|+C \\ & =x+4 \sqrt{x}+4 \ln |\sqrt{x}-1|+C \end{aligned}